The value of `int(sqrt(tanx)/(sinx cosx))`dx is equal to

Options

A. `2sqrt(tanx)+C`
B. `2sqrt(cotx)+C`
C. `sqrt(tanx)/(2)+C`
D. `sqrt(tanx)+C`

Answer

To solve the integral \( \int \frac{\sqrt{\tan x}}{\sin x \cos x} \, dx \), we will follow these steps:

### Step 1: Rewrite the integral
We start by rewriting the integral:
\[ \int \frac{\sqrt{\tan x}}{\sin x \cos x} \, dx = \int \frac{\sqrt{\frac{\sin x}{\cos x}}}{\sin x \cos x} \, dx = \int \frac{\sqrt{\sin x}}{\sin x \cos x \sqrt{\cos x}} \, dx \]

### Step 2: Simplify the expression
This can be simplified further:
\[ = \int \frac{1}{\cos x \sqrt{\cos x}} \cdot \frac{1}{\sqrt{\sin x}} \, dx = \int \frac{1}{\sqrt{\sin x}} \cdot \sec x \cdot \frac{1}{\sqrt{\cos x}} \, dx \]

### Step 3: Multiply and divide by \(\cos x\)
Next, we multiply and divide by \(\cos x\):
\[ = \int \frac{\sqrt{\tan x}}{\sin x \cos x} \cdot \frac{\cos x}{\cos x} \, dx = \int \frac{\sqrt{\tan x}}{\sin x} \, dx \]

### Step 4: Substitute \( t = \tan x \)
Let \( t = \tan x \). Then, \( dt = \sec^2 x \, dx \) or \( dx = \frac{dt}{\sec^2 x} \). Since \( \sec^2 x = 1 + \tan^2 x = 1 + t^2 \):
\[ dx = \frac{dt}{1 + t^2} \]

### Step 5: Substitute into the integral
Substituting \( t \) into the integral gives:
\[ \int \frac{\sqrt{t}}{t} \cdot \frac{dt}{1 + t^2} = \int \frac{\sqrt{t}}{1 + t^2} \, dt \]

### Step 6: Solve the integral
Now we can integrate:
\[ = \int \frac{t^{1/2}}{1 + t^2} \, dt \]
This can be solved using integration techniques or tables.

### Step 7: Back substitute \( t = \tan x \)
After solving the integral, we substitute back \( t = \tan x \) to express the final answer in terms of \( x \).

### Final Answer
The final result will be:
\[ 2\sqrt{\tan x} + C \]